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Question

# At 800oC, the following equilibrium is established as:F2(g)⇌2F(g)The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pinhole. It is found that at 800oC and 1 atm mixture effuses 1.6 times as fast as SO2 effuses under the similar conditions. (At.wt. of F=19). What is the value of Kp (in atm)?

A
0.315
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B
0.685
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C
0.46
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D
1.49
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Solution

## The correct option is D 1.49The rate of effusion is inversely proportional to square of the molar mass.rmixrSO2=√MSO2MmixHere, rmix and rSO2 are the rates of effusion of mixture and sulphur dioxide respectively. MSO2 and Mmix are the molar masses of SO2 and mixture respectively.Substitute values in the above expression-1.6=√64Mmix2.56=64MmixM=25Hence, the molecular weight of the mixture is 25 g/mol respectively. Let x be the mole fraction of F2 in the mixture. The mole fraction of F will be 1−x.The molar mass of the mixture is 38x+19−19x=25 19x=6The mole fraction of F2 is x=0.32The mole fraction of F is1−x=0.68The partial pressure is the product of mole fraction and total pressure which is 1 atm.The equilibrium constant expression is Kp=P2FPF2=(0.68)20.32=1.49

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